If $0<\mu(X)<\infty$ and $0<p<q< \infty$. Is $(\int_X |f|^p) ^\frac{1}{p}\leq \int_X |f|^q d\mu) ^\frac{1}{q}$ true?

Question

If $0<\mu(X)<\infty$ and $0<p<q< \infty$ Is $(\int_X |f|^p d\mu) ^\frac{1}{p}\leq \int_X |f|^q d\mu) ^\frac{1}{q}$ true.

I should say that $\frac{1}{p}+\frac{1}{q}=1$

I think it is but cannot seem to prove, and it is not in my notes. Any help would be appreciated. I cannot seem to get started, I apologize.

Answer 1

Since $\frac{p}{q}+\frac{q-p}{q}=1$, apply Holder's inequality.

$$ \left(\int_X |f|^p\,d\mu\right)^{1/p}\leq \left[\left(\int_X |f|^{q}\right)^{p/q}\mu(X)^{(q-p)/q}\right]^{1/p}=\left(\int_X|f|^q\right)^{1/q}\mu(X)^{(q-p)/(pq)}. $$

Note Holder's inequality only applies if the condition $\mu(X)<\infty$ holds.

The inequality you ask for will hold provided $\mu(X)\leq1$.